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I'm not sure if this is a true statement or not, but due to the compactness, it seems like it should be. My attempt at a proof involves supposing it has infinitely many components, and then taking a sequence such that each element in the sequence is in a unique component. Since compactness implies limit-pint compactness, we know this infinite subset of has a limit point. From here, it seems like this limit point should provide a contradiction to the construction, but I haven't been able to show any contradiction yet...

Essentially what I want to do is show that a component is closed and open, and if there are only finitely many, we get this for free.

Note that this is part of a larger problem, this is just what the guts of my proof comes down to, so if it isn't true, the direction of my proof is entirely wrong...

Thanks in advance!!

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By the way, we know that components in a compact, Hausdorff space are disjoint since components and quasicomponents agree. Thus, my sequence isn't too ambiguous... –  notsure Oct 25 '12 at 15:53
    
Note the compactness doesn't imply sequential compactness in general. See math.stackexchange.com/questions/220422/… –  Jason DeVito Oct 25 '12 at 15:56
    
That is certainly true, I wasn't suggesting limit-point compactness was equivalent to sequentially compact, merely that there had to exists a limit point of the sequence. –  notsure Oct 25 '12 at 15:59

2 Answers 2

No, it’s not true. The middle-thirds Cantor set is totally disconnected, meaning that its components are all singletons. Since it has cardinality $2^\omega=\mathfrak c$, the same cardinality as $\Bbb R$, it has far more than finitely many components.

In fact it has an even stronger property: it’s zero-dimensional (in the sense of small inductive dimension, which is the usual sense when the term is used without further qualification), meaning that it has a base of clopen sets, and it’s $T_1$; every zero-dimensional $T_1$-space is totally disconnected, but not conversely.

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Thanks! I should have thought about that, it just didn't appear to me... Thanks again! –  notsure Oct 25 '12 at 16:01
    
@notsure: You’re welcome! –  Brian M. Scott Oct 25 '12 at 16:04

For a simpler example, note that $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\ldots\}$ (with its natural topology as a subspace of $\Bbb{R}$ is compact, Hausdorff, and has $\aleph_0$ components.

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